In the following four periods
(i) Time of revolution of a satellite just above the earth’s surface $(T_{s t})$
(ii) Period of oscillation of mass inside the tunnel bored along the diameter of the earth $(T_{m a})$
(iii) Period of simple pendulum having a length equal to the earth’s radius in a uniform field of 9.8 N/k $(T_{s p})$
(iv) Period of an infinite length simple pendulum in the earth’s real gravitational field $(T_{i s})$

T_{s t} > T_{m a}

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T_{m a} > T_{s t}

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T_{s p} < T_{i s}

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T_{s t} = T_{m a} = T_{s p} = T_{i s}

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Solution

The correct option is C $T_{s p} < T_{i s}$
$(i) T_{s t} = 2 π \sqrt{\frac{(R + h)^{3}}{G M}} = 2 π \sqrt{\frac{R}{g}} [A s h << R a n d G M = R^{2}] (i i) T_{m a} = 2 π \sqrt{\frac{R}{g}} (i i i) T_{s p} = 2 π \sqrt{\frac{1}{g (\frac{1}{l} + \frac{1}{R})}} = 2 π \sqrt{\frac{R}{2 g}} [A s l = R] (i v) T_{i s} = 2 π \sqrt{\frac{R}{g}} [A s l = \infty]$

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Similar questions

Q. In the following four periods
(i) Time of revolution of a satellite just above the earth’s surface

(T_{s t})

(ii) Period of oscillation of mass inside the tunnel bored along the diameter of the earth

(T_{m a})

(iii) Period of simple pendulum having a length equal to the earth’s radius in a uniform field of 9.8 N/k

(T_{s p})

(iv) Period of an infinite length simple pendulum in the earth’s real gravitational field

(T_{i s})

Q. A simple pendulum of length 1 m is taken to height h (h = R = radius of the earth) from the earth’s surface.
The time period of small oscillation of the pendulum at that point is

(g = π^{2} m s^{- 2})

Q. Time period of a satellite revolving above Earth’s surface at a height equal to R, where R the radius of Earth, is

Q. Support of a simple pendulum of length

l

is being accelerated in a plane at an angle of

30^{\circ}

with the horizontal as shown in the diagram with an acceleration of magnitude

g

(earth’s acceleration due to gravity). Find the new time period of the oscillation of simple pendulum.

Periodic time of a satellite revolving above Earth’s surface at a height equal to R, where R the radius of Earth, is [g is acceleration due to gravity at Earth’s surface]

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The correct option is C Tsp<Tis(i) Tst=2π√(R+h)3GM=2π√Rg [As h<<R and GM=R2](ii) Tma=2π√Rg(iii) Tsp=2π√1g(1l+1R)=2π√R2g [As l=R](iv) Tis=2π√Rg [As l=∞]